Question

1-Simplifique as frações algébricas
$\frac{8c}{4a} =$
$\frac{4c {}^{2} }{6c {}^{9} } =$
$\frac{abc}{c {}^{9} } =$
$\frac{7ac {}^{9} }{21ac {}^{5} } =$
2-Utilize a fatoraçao da diferença de dois quadrados ou a fatoraçao do trinômio quadrado perfeito e simplifique:
$\frac{7 \times - 7y}{5x {}^{2} - 5y {}^{2} } =$
$\frac{5x {}^{2} - 5}{4x + 4} =$
$\frac{x {}^{2} - 9 }{x {}^{2} + 3x} =$

Answer 1

$a)\\\\\frac{8c}{4a} =\frac{2c}{a} \\ \\ b)\\ \\ \frac{4c^{2}}{6c^{9}} =\frac{2}{3c^{7}} \\ \\ c)\\ \\ \frac{abc}{c^{9}} =\frac{ab}{c^{8}} \\ \\ d)\\ \\ \frac{7ac^{9}}{21ac^{5}} =\frac{c^{4}}{3}$

$2.a)\\ \\ \frac{7x-7y}{5x^{2}-5y^{2}} =\frac{7(x-y)}{5(x^{2}-y^{2})} =\frac{7(x-y)}{5(x+y)(x-y)} =\frac{7}{5x+5y}$

$2.b)\\ \\ \frac{5x^{2}-5}{4x+4} =\frac{5(x^{2}-1)}{4(x+1)} =\frac{5(x+1)(x-1)}{4(x+1)} =\frac{5x-5}{4}$

$\frac{x^{2}-9}{x^{2}+3x} =\frac{(x+3)(x-3)}{x(x+3)} =\frac{x-3}{x}$

Answer 2

= 8c/4a
= 8/4 . c/a
= 2c/a

= 4c^2 / 6c^9
= 4/6 . c^2 / c^9
= 2/3 . 1/c^7
= 2/(3c^7)

= abc /c^9
= ab/(c^8)

= 7ac^9 / 21ac^5
= 7/21 . ac^9 / ac^5
= 1/3 . c^4
= c^4 / 3

2)

= (7x - 7y) / (5x^2 - 5y^2)
= 7.(x - y) / 5.(x^2 - y^2)
= 7.(x - y)/ 5.(x - y).(x + y)
= 7/[5.(x + y)]

= ( 5x^2 - 5) / (4x + 4)
= 5.(x^2 - 1) / 4.(x + 1)
= 5.(x + 1).(x - 1) / 4.(x + 1)
= 5(x - 1)/4

= (x^2 - 9) / (x^2 + 3x)
= (x + 3).(x - 3) / x.(x + 3)
= (x - 3)/ x